Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{2x + 4}{x + 9} \div \dfrac{x^2 + 11x + 18}{x^2 + 9x} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{2x + 4}{x + 9} \times \dfrac{x^2 + 9x}{x^2 + 11x + 18} $ First factor the quadratic. $r = \dfrac{2x + 4}{x + 9} \times \dfrac{x^2 + 9x}{(x + 2)(x + 9)} $ Then factor out any other terms. $r = \dfrac{2(x + 2)}{x + 9} \times \dfrac{x(x + 9)}{(x + 2)(x + 9)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 2(x + 2) \times x(x + 9) } { (x + 9) \times (x + 2)(x + 9) } $ $r = \dfrac{ 2x(x + 2)(x + 9)}{ (x + 9)(x + 2)(x + 9)} $ Notice that $(x + 9)$ and $(x + 2)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 2x\cancel{(x + 2)}(x + 9)}{ (x + 9)\cancel{(x + 2)}(x + 9)} $ We are dividing by $x + 2$ , so $x + 2 \neq 0$ Therefore, $x \neq -2$ $r = \dfrac{ 2x\cancel{(x + 2)}\cancel{(x + 9)}}{ \cancel{(x + 9)}\cancel{(x + 2)}(x + 9)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $r = \dfrac{2x}{x + 9} ; \space x \neq -2 ; \space x \neq -9 $